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Setler79 [48]
2 years ago
11

Mrs. Lovejoy and Christi spent the day baking cookies. When they divided them into 8 stacks, they had 14 cookies in each stack.

How many cookies did they have? Solve and label the word problem. Please I need help.​
Mathematics
1 answer:
Rasek [7]2 years ago
7 0

Answer:

they had 112 cookies

Step-by-step explanation:

14×8 = 112

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For which equation is m = 12 the solution? a. 4m=40 b. m+20=42. c. 4m=48. d. m-4=9
Gelneren [198K]
To solve this question, we just need to insert 12 into the m position of each question and see if the equation holds true.

a. 4m = 40

4(12) = 40

48 = 40

48 obviously does not equal 40, so it is not choice A.

b. m + 20 = 42

12 + 20 = 42

32 = 42

Again, 32 isn't the same as 42, so not choice B either.

c. 4m = 48

4(12) = 48

48 = 48

It looks like this one is true, but let's solve D also just to make sure.

d. m - 4 = 9

12 - 4 = 9

8 = 9

This is false, since 8 does not equal 9.

Therefore, choice C (4m = 48) is the correct answer.
Hope that helped! =)
6 0
3 years ago
Read 2 more answers
Ok this has nothing to so with my assignments or anything but i have a serious question please help so my question is....
umka2103 [35]
If she is stated in the will she can. If not, then she can't unless You simply give it to her once you inherit the money.
5 0
3 years ago
32. Which of the following is horizontal ellipse
maria [59]

Answer:

  • b. 16x² + 25y² = 1

Step-by-step explanation:

Only one of them has a < b in ax² + by² = c formula, making its graph horizontal

All the rest are vertical, see attached

6 0
3 years ago
Hari bought a shirt for Rs 320 and sold it for Rs 365 to Ram calculate his profit or loss​
solmaris [256]

Answer:

profit=sp-cp

=365-320

=45#

7 0
3 years ago
1-cot^2a+cot^4a=sin^2a(1+cot^6a) prove it.​
aliina [53]

Step-by-step explanation:

We have

1-cot²a + cot⁴a = sin²a(1+cot⁶a)

First, we can take a look at the right side. It expands to sin²a + cot⁶(a)sin²(a) = sin²a + cos⁶a/sin⁴a (this is the expanded right side) as cot(a) = cos(a)/sin(a), so cos⁶a = cos⁶a/sin⁶a. Therefore, it might be helpful to put everything in terms of sine and cosine to solve this.

We know 1 = sin²a+cos²a and cot(a) = cos(a)/sin(a), so we have

1-cot²a + cot⁴a = sin²a+cos²a-cos²a/sin²a + cos⁴a/sin⁴a

Next, we know that in the expanded right side, we have sin²a + something. We can use that to isolate the sin²a. The rest of the expanded right side has a denominator of sin⁴a, so we can make everything else have that denominator.

sin²a+cos²a-cos²a/sin²a + cos⁴a/sin⁴a

= sin²a + (cos²(a)sin⁴(a) - cos²(a)sin²(a) + cos⁴a)/sin⁴a

We can then factor cos²a out of the numerator

sin²a + (cos²(a)sin⁴(a) - cos²(a)sin²(a) + cos⁴a)/sin⁴a

= sin²a + cos²a (sin⁴a-sin²a+cos²a)/sin⁴a

Then, in the expanded right side, we can notice that the fraction has a numerator with only cos in it. We can therefore write sin⁴a in terms of cos (we don't want to write the sin²a term in terms of cos because it can easily add with cos²a to become 1, so we can hold that off for later) , so

sin²a = (1-cos²a)

sin⁴a = (1-cos²a)² = cos⁴a - 2cos²a + 1

sin²a + cos²a (sin⁴a-sin²a+cos²a)/sin⁴a

= sin²a + cos²a (cos⁴a-2cos²a+1-sin²a+cos²a)/sin⁴a

= sin²a + cos²a (cos⁴a-cos²a+1-sin²a)/sin⁴a

factor our the -cos²a-sin²a as -1(cos²a+sin²a) = -1(1) = -1

sin²a + cos²a (cos⁴a-cos²a+1-sin²a)/sin⁴a

=  sin²a + cos²a (cos⁴a-1 + 1)/sin⁴a

= sin²a + cos⁶a/sin⁴a

= sin²a(1+cos⁶a/sin⁶a)

= sin²a(1+cot⁶a)

8 0
3 years ago
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